Integrand size = 25, antiderivative size = 357 \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx=-\frac {(A-i B) x}{4 (a-i b)^{2/3}}-\frac {(A+i B) x}{4 (a+i b)^{2/3}}-\frac {\sqrt {3} (i A+B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 (a-i b)^{2/3} d}+\frac {\sqrt {3} (i A-B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 (a+i b)^{2/3} d}-\frac {(i A-B) \log (\cos (c+d x))}{4 (a+i b)^{2/3} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 (a-i b)^{2/3} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{2/3} d}-\frac {3 (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{2/3} d} \]
-1/4*(A-I*B)*x/(a-I*b)^(2/3)-1/4*(A+I*B)*x/(a+I*b)^(2/3)-1/4*(I*A-B)*ln(co s(d*x+c))/(a+I*b)^(2/3)/d+1/4*(I*A+B)*ln(cos(d*x+c))/(a-I*b)^(2/3)/d+3/4*( I*A+B)*ln((a-I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/(a-I*b)^(2/3)/d-3/4*(I*A-B )*ln((a+I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/(a+I*b)^(2/3)/d-1/2*(I*A+B)*arc tan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a-I*b)^(1/3))*3^(1/2))*3^(1/2)/(a-I*b )^(2/3)/d+1/2*(I*A-B)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a+I*b)^(1/3) )*3^(1/2))*3^(1/2)/(a+I*b)^(2/3)/d
Time = 0.27 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.85 \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx=\frac {i \left (-\frac {(A-i B) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )+\log \left ((a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )\right )}{(a-i b)^{2/3}}+\frac {(A+i B) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )+\log \left ((a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}\right )\right )}{(a+i b)^{2/3}}\right )}{4 d} \]
((I/4)*(-(((A - I*B)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3)) /(a - I*b)^(1/3))/Sqrt[3]] - 2*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^ (1/3)] + Log[(a - I*b)^(2/3) + (a - I*b)^(1/3)*(a + b*Tan[c + d*x])^(1/3) + (a + b*Tan[c + d*x])^(2/3)]))/(a - I*b)^(2/3)) + ((A + I*B)*(2*Sqrt[3]*A rcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] - 2*Lo g[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] + Log[(a + I*b)^(2/3) + (a + I*b)^(1/3)*(a + b*Tan[c + d*x])^(1/3) + (a + b*Tan[c + d*x])^(2/3)]))/( a + I*b)^(2/3)))/d
Time = 0.57 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.65, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4022, 3042, 4020, 25, 69, 16, 1082, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}}dx\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {1}{2} (A+i B) \int \frac {1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}}dx+\frac {1}{2} (A-i B) \int \frac {i \tan (c+d x)+1}{(a+b \tan (c+d x))^{2/3}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (A+i B) \int \frac {1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}}dx+\frac {1}{2} (A-i B) \int \frac {i \tan (c+d x)+1}{(a+b \tan (c+d x))^{2/3}}dx\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) (a+b \tan (c+d x))^{2/3}}d(i \tan (c+d x))}{2 d}-\frac {i (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) (a+b \tan (c+d x))^{2/3}}d(-i \tan (c+d x))}{2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {i (A+i B) \int \frac {1}{(i \tan (c+d x)+1) (a+b \tan (c+d x))^{2/3}}d(-i \tan (c+d x))}{2 d}-\frac {i (A-i B) \int \frac {1}{(1-i \tan (c+d x)) (a+b \tan (c+d x))^{2/3}}d(i \tan (c+d x))}{2 d}\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {i (A-i B) \left (-\frac {3 \int \frac {1}{-\tan ^2(c+d x)+(a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a-i b}}-\frac {3 \int \frac {1}{\sqrt [3]{a-i b}-i \tan (c+d x)}d\sqrt [3]{a+b \tan (c+d x)}}{2 (a-i b)^{2/3}}-\frac {\log (1-i \tan (c+d x))}{2 (a-i b)^{2/3}}\right )}{2 d}-\frac {i (A+i B) \left (-\frac {3 \int \frac {1}{-\tan ^2(c+d x)+(a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a+i b}}-\frac {3 \int \frac {1}{i \tan (c+d x)+\sqrt [3]{a+i b}}d\sqrt [3]{a+b \tan (c+d x)}}{2 (a+i b)^{2/3}}-\frac {\log (1+i \tan (c+d x))}{2 (a+i b)^{2/3}}\right )}{2 d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {i (A-i B) \left (-\frac {3 \int \frac {1}{-\tan ^2(c+d x)+(a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a-i b}}-\frac {\log (1-i \tan (c+d x))}{2 (a-i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 (a-i b)^{2/3}}\right )}{2 d}-\frac {i (A+i B) \left (-\frac {3 \int \frac {1}{-\tan ^2(c+d x)+(a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a+i b}}-\frac {\log (1+i \tan (c+d x))}{2 (a+i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 (a+i b)^{2/3}}\right )}{2 d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {i (A-i B) \left (\frac {3 \int \frac {1}{\tan ^2(c+d x)-3}d\left (\frac {2 i \tan (c+d x)}{\sqrt [3]{a-i b}}+1\right )}{(a-i b)^{2/3}}-\frac {\log (1-i \tan (c+d x))}{2 (a-i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 (a-i b)^{2/3}}\right )}{2 d}-\frac {i (A+i B) \left (\frac {3 \int \frac {1}{\tan ^2(c+d x)-3}d\left (1-\frac {2 i \tan (c+d x)}{\sqrt [3]{a+i b}}\right )}{(a+i b)^{2/3}}-\frac {\log (1+i \tan (c+d x))}{2 (a+i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 (a+i b)^{2/3}}\right )}{2 d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {i (A-i B) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (c+d x)}{\sqrt {3}}\right )}{(a-i b)^{2/3}}-\frac {\log (1-i \tan (c+d x))}{2 (a-i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 (a-i b)^{2/3}}\right )}{2 d}-\frac {i (A+i B) \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (c+d x)}{\sqrt {3}}\right )}{(a+i b)^{2/3}}-\frac {\log (1+i \tan (c+d x))}{2 (a+i b)^{2/3}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 (a+i b)^{2/3}}\right )}{2 d}\) |
((I/2)*(A - I*B)*(((-I)*Sqrt[3]*ArcTanh[Tan[c + d*x]/Sqrt[3]])/(a - I*b)^( 2/3) - Log[1 - I*Tan[c + d*x]]/(2*(a - I*b)^(2/3)) + (3*Log[(a - I*b)^(1/3 ) - I*Tan[c + d*x]])/(2*(a - I*b)^(2/3))))/d - ((I/2)*(A + I*B)*((I*Sqrt[3 ]*ArcTanh[Tan[c + d*x]/Sqrt[3]])/(a + I*b)^(2/3) - Log[1 + I*Tan[c + d*x]] /(2*(a + I*b)^(2/3)) + (3*Log[(a + I*b)^(1/3) + I*Tan[c + d*x]])/(2*(a + I *b)^(2/3))))/d
3.5.76.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Simp[3/(2*b*q) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 /3)], x] - Simp[3/(2*b*q^2) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.40 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.19
| method | result | size |
| derivativedivides | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (B \,\textit {\_R}^{3}+A b -B a \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 d}\) | \(69\) |
| default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (B \,\textit {\_R}^{3}+A b -B a \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 d}\) | \(69\) |
| parts | \(\frac {A b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{2} \left (\textit {\_R}^{3}-a \right )}\right )}{2 d}+\frac {B \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{2 d}\) | \(106\) |
1/2/d*sum((B*_R^3+A*b-B*a)/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R= RootOf(_Z^6-2*_Z^3*a+a^2+b^2))
Leaf count of result is larger than twice the leaf count of optimal. 6073 vs. \(2 (263) = 526\).
Time = 1.35 (sec) , antiderivative size = 6073, normalized size of antiderivative = 17.01 \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx=\text {Too large to display} \]
\[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
Time = 23.04 (sec) , antiderivative size = 4562, normalized size of antiderivative = 12.78 \[ \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx=\text {Too large to display} \]
log((((16*(-B^6*a^2*b^2*d^6)^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*( a^2 + b^2)^2))^(1/3)*(1944*a*b^4*(-B^6*a^2*b^2*d^6)^(1/2) - 1944*B^3*a*b^6 *d^3 + 243*B*b^8*d^5*(a + b*tan(c + d*x))^(1/3)*((16*(-B^6*a^2*b^2*d^6)^(1 /2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3) - 243*B*a^ 4*b^4*d^5*(a + b*tan(c + d*x))^(1/3)*((16*(-B^6*a^2*b^2*d^6)^(1/2) + 8*B^3 *a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3)))/(4*d^6*(a^2 + b^2)) + (486*B^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*((((16*B^3*a^2*d^3 - 16*B ^3*b^2*d^3)^2/4 - B^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2*b^2*d^6)))^ (1/3) + log((486*B^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 - ((-(16*(-B^6*a^ 2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(1/ 3)*(1944*a*b^4*(-B^6*a^2*b^2*d^6)^(1/2) + 1944*B^3*a*b^6*d^3 - 243*B*b^8*d ^5*(a + b*tan(c + d*x))^(1/3)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d ^3 + 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3) + 243*B*a^4*b^4*d^5*(a + b* tan(c + d*x))^(1/3)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3 *b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3)))/(4*d^6*(a^2 + b^2)))*(-(((16*B^3*a^ 2*d^3 - 16*B^3*b^2*d^3)^2/4 - B^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d ^6))^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2 *b^2*d^6)))^(1/3) + log(((((1944*a*b^4*(a^2 + b^2)*((8*(-A^6*d^6*(a^2 - b^ 2)^2)^(1/2) + 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3) + (7776*A*a*b^...